/*
4Sum II
=======

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.

To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.

Example:
Input:
A = [ 1, 2]
B = [-2,-1]
C = [-1, 2]
D = [ 0, 2]
Output:
2

Explanation:
The two tuples are:
1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0
*/

class Solution
{
public:
  int fourSumCount(vector<int> &A, vector<int> &B, vector<int> &C, vector<int> &D)
  {
    unordered_map<int, int> abSum;
    for (auto a : A)
    {
      for (auto b : B)
      {
        ++abSum[a + b];
      }
    }
    int count = 0;
    for (auto c : C)
    {
      for (auto d : D)
      {
        auto it = abSum.find(0 - c - d);
        if (it != abSum.end())
        {
          count += it->second;
        }
      }
    }
    return count;
  }
};
